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Secondary 4 Pure Physics Preliminary Examination Paper 2

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Questions

TuitionGoWhere Practice Paper - Pure Physics Secondary 4

TuitionGoWhere Secondary School (AI)
Subject: Pure Physics
Level: Secondary 4
Paper: PRELIM (Version 2)
Duration: 1 hour 45 minutes
Total Marks: 80

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided above.
Answer all questions.
Write your answers in the spaces provided on the question paper.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a calculator.
Where appropriate, take $g = 10 \text{ N/kg}$ .
The total number of marks for this paper is 80.

Section A: Multiple Choice Questions [20 marks]

Answer all questions. For each question, choose the correct option and write the letter (A, B, C, or D) in the box provided.

1

A transformer has 400 turns on its primary coil and 100 turns on its secondary coil. The primary coil is connected to a 240 V a.c. supply. What is the voltage across the secondary coil?
[1]

☐ A 60 V
☐ B 120 V
☐ C 480 V
☐ D 960 V

2

The diagram below shows a current-carrying wire placed between the poles of a magnet. The current flows into the page.

<image_placeholder> id: Q2-fig1 type: diagram linked_question: Q2 description: A uniform magnetic field directed from North (left) to South (right). A wire carries current into the page (represented by a cross). Label the North and South poles, magnetic field direction (left to right), and current direction (into page). labels: N (left), S (right), B-field arrow left to right, current symbol (cross) at wire values: None must_show: Uniform magnetic field lines, wire with current into page, N and S poles labelled </image_placeholder>

What is the direction of the force acting on the wire?
[1]

☐ A Upwards
☐ B Downwards
☐ C To the left
☐ D To the right

3

An electric kettle rated 2.2 kW, 240 V is used for 15 minutes. What is the electrical energy consumed?
[1]

☐ A 0.55 kWh
☐ B 1.10 kWh
☐ C 33 kWh
☐ D 55 kWh

4

A copper wire of length 2.0 m and cross-sectional area $1.0 \times 10^{-6} \text{ m}^2$ has a resistance of 0.34 $\Omega$ . What is the resistivity of copper?
[1]

☐ A $1.7 \times 10^{-7} \ \Omega \text{m}$
☐ B $1.7 \times 10^{-6} \ \Omega \text{m}$
☐ C $6.8 \times 10^{-7} \ \Omega \text{m}$
☐ D $6.8 \times 10^{-6} \ \Omega \text{m}$

5

Which of the following statements about electromagnetic induction is correct?
[1]

☐ A An induced e.m.f. is produced only when a magnet moves towards a coil.
☐ B The magnitude of the induced e.m.f. is independent of the rate of change of magnetic flux.
☐ C The direction of the induced current opposes the change producing it.
☐ D A stationary magnet inside a coil produces a steady induced current.

6

A circuit consists of a 12 V battery, a 4 $\Omega$ resistor, and a 2 $\Omega$ resistor connected in series. What is the potential difference across the 2 $\Omega$ resistor?
[1]

☐ A 2 V
☐ B 4 V
☐ C 6 V
☐ D 8 V

7

The diagram shows a simple a.c. generator. The coil rotates in a uniform magnetic field.

<image_placeholder> id: Q7-fig1 type: diagram linked_question: Q7 description: A rectangular coil rotating in a uniform magnetic field between N and S poles. Slip rings and brushes connected to external circuit. Show coil at horizontal position (parallel to field). labels: N, S, magnetic field lines (N to S), coil sides AB and CD, slip rings, brushes, rotation arrow values: None must_show: Coil horizontal (parallel to B-field), slip rings, brushes, N/S poles, field direction </image_placeholder>

At the instant shown, the coil is horizontal. Which statement describes the induced e.m.f. at this instant?
[1]

☐ A The induced e.m.f. is zero.
☐ B The induced e.m.f. is at its maximum value.
☐ C The induced e.m.f. is increasing from zero.
☐ D The induced e.m.f. is decreasing from maximum.

8

A household circuit has a 13 A fuse. The mains voltage is 240 V. What is the maximum number of 60 W lamps that can be connected in parallel on this circuit?
[1]

☐ A 26
☐ B 32
☐ C 48
☐ D 52

9

A positively charged rod is brought near an uncharged metal sphere. The sphere is then earthed momentarily while the rod is still near. The rod is then removed. What is the final charge on the sphere?
[1]

☐ A Positive
☐ B Negative
☐ C Neutral
☐ D Cannot be determined

10

The graph below shows the current-voltage characteristic of a filament lamp.

<image_placeholder> id: Q10-fig1 type: graph linked_question: Q10 description: I-V graph for a filament lamp. Current on y-axis, Voltage on x-axis. Curve starts steep and gradually bends over (gradient decreases) as voltage increases. labels: I (A) on y-axis, V (V) on x-axis values: Curve passes through origin, gradient decreases with increasing V must_show: Non-linear curve through origin with decreasing gradient </image_placeholder>

Which statement explains the shape of the graph?
[1]

☐ A The resistance of the filament decreases as temperature increases.
☐ B The resistance of the filament increases as temperature increases.
☐ C The filament obeys Ohm's Law at all voltages.
☐ D The filament has zero resistance at low voltages.

11

A step-down transformer has 800 turns on the primary coil and 200 turns on the secondary coil. The primary current is 0.5 A. Assuming 100% efficiency, what is the secondary current?
[1]

☐ A 0.125 A
☐ B 0.5 A
☐ C 2.0 A
☐ D 4.0 A

12

The diagram shows a cathode-ray oscilloscope (CRO) trace for an a.c. voltage signal. The time-base is set to 5 ms/div and the Y-gain is set to 2 V/div.

<image_placeholder> id: Q12-fig1 type: diagram linked_question: Q12 description: CRO screen showing a sinusoidal wave. 2 complete cycles span 4 horizontal divisions. Peak-to-peak vertical displacement is 3 divisions. labels: Time-base: 5 ms/div, Y-gain: 2 V/div values: 4 divisions per 2 cycles horizontally, 3 divisions peak-to-peak vertically must_show: Sinusoidal wave, grid divisions visible, 2 cycles across 4 divisions, amplitude 1.5 divisions </image_placeholder>

What is the frequency of the a.c. signal?
[1]

☐ A 25 Hz
☐ B 50 Hz
☐ C 100 Hz
☐ D 200 Hz

13

A wire of resistance $R$ is stretched uniformly until its length doubles. Its volume remains constant. What is the new resistance of the wire?
[1]

☐ A $R$
☐ B $2R$
☐ C $4R$
☐ D $8R$

14

In a household wiring circuit, the live wire is brown, the neutral wire is blue, and the earth wire is green/yellow. Which wire is connected to the fuse in a 3-pin plug?
[1]

☐ A Live wire
☐ B Neutral wire
☐ C Earth wire
☐ D Both live and neutral wires

15

A bar magnet is dropped through a long copper tube. It falls slower than a non-magnetic object of the same mass and size. Which principle explains this?
[1]

☐ A Faraday's Law of Electromagnetic Induction
☐ B Lenz's Law
☐ C Ampere's Law
☐ D Coulomb's Law

16

The diagram shows a potential divider circuit. The battery has e.m.f. 12 V and negligible internal resistance. The variable resistor is set to 6 $\Omega$ and the fixed resistor is 3 $\Omega$ .

<image_placeholder> id: Q16-fig1 type: diagram linked_question: Q16 description: Potential divider circuit: 12 V battery connected in series with a variable resistor (6 Ω) and a fixed resistor (3 Ω). Voltmeter connected across the fixed resistor. labels: 12 V battery, variable resistor 6 Ω, fixed resistor 3 Ω, voltmeter across fixed resistor values: 12 V, 6 Ω, 3 Ω must_show: Series circuit with voltmeter across 3 Ω resistor </image_placeholder>

What is the reading on the voltmeter?
[1]

☐ A 3 V
☐ B 4 V
☐ C 6 V
☐ D 8 V

17

An electron enters a uniform magnetic field at right angles to the field lines. The electron moves in a circular path. What provides the centripetal force?
[1]

☐ A Electric force
☐ B Magnetic force
☐ C Gravitational force
☐ D Frictional force

18

A 100 $\Omega$ resistor and a 200 $\Omega$ resistor are connected in parallel across a 12 V battery. What is the total power dissipated in the circuit?
[1]

☐ A 0.72 W
☐ B 1.44 W
☐ C 2.16 W
☐ D 2.88 W

19

The diagram shows a magnetic field pattern around a current-carrying solenoid.

<image_placeholder> id: Q19-fig1 type: diagram linked_question: Q19 description: Solenoid with current direction shown. Magnetic field lines emerge from one end (N) and enter the other (S). Field lines inside are parallel and uniform. labels: Current direction arrows on coils, N and S poles, field lines inside and outside values: None must_show: Uniform field inside solenoid, field lines emerging from N and entering S, current direction </image_placeholder>

Which end of the solenoid is the North pole?
[1]

☐ A End P
☐ B End Q
☐ C Both ends
☐ D Neither end

20

A capacitor is charged through a resistor from a constant voltage source. Which graph shows how the voltage across the capacitor varies with time?

<image_placeholder> id: Q20-fig1 type: graph linked_question: Q20 description: Four graph options showing V vs t. Correct graph: exponential rise from 0, asymptotically approaching V_max. labels: V (V) on y-axis, t (s) on x-axis values: V_max labelled on y-axis must_show: Exponential charging curve starting at origin, asymptote at V_max </image_placeholder>

[1]

Section B: Structured Questions [45 marks]

Answer all questions in the spaces provided.

21

A student sets up a circuit to investigate the relationship between current and voltage for a metallic conductor at constant temperature. The circuit includes a battery, a variable resistor, an ammeter, a voltmeter, and the conductor.

(a) Draw the circuit diagram using standard circuit symbols. Label all components.
[3]

(b) The student obtains the following data:

Voltage / V	0.5	1.0	1.5	2.0	2.5	3.0
Current / A	0.2	0.4	0.6	0.8	1.0	1.2

Plot the graph of current against voltage on the grid below.

<image_placeholder> id: Q21b-fig1 type: graph linked_question: Q21 description: Blank grid for I-V graph. x-axis: Voltage (V) from 0 to 3.5 V. y-axis: Current (A) from 0 to 1.4 A. Grid lines at 0.5 V and 0.2 A intervals. labels: Voltage / V on x-axis, Current / A on y-axis values: Scale: 0-3.5 V, 0-1.4 A must_show: Labelled axes with units, appropriate scale, grid lines </image_placeholder>

[3]

(c) From the graph, determine the resistance of the conductor.
[2]

(d) State the law that describes the relationship between current and voltage for this conductor.
[1]

22

A transformer is used to step down the voltage from 240 V to 12 V for a low-voltage lighting system. The primary coil has 1200 turns. The secondary coil supplies a current of 4.0 A to the lamps. The transformer is 90% efficient.

(a) Calculate the number of turns on the secondary coil.
[2]

(b) Calculate the current in the primary coil.
[3]

(c) Explain why the transformer core is laminated.
[2]

(d) State one reason why the transformer is not 100% efficient.
[1]

23

The diagram shows a simple d.c. motor. The coil ABCD is placed in a uniform magnetic field. Current flows from A to B to C to D.

<image_placeholder> id: Q23-fig1 type: diagram linked_question: Q23 description: Rectangular coil ABCD in uniform magnetic field (N left, S right). Current direction: A→B→C→D. Commutator and brushes shown. Coil is horizontal (AB top, CD bottom). labels: N, S, magnetic field lines (left to right), coil sides AB, BC, CD, DA, current arrows, commutator, brushes values: None must_show: Coil horizontal, current direction, N/S poles, field direction, commutator </image_placeholder>

(a) On the diagram, draw arrows to show the direction of the force acting on side AB and side CD of the coil.
[2]

(b) Explain why the coil experiences a turning effect (torque).
[2]

(c) State the function of the commutator in a d.c. motor.
[1]

(d) The coil has 50 turns, each of area $4.0 \times 10^{-3} \text{ m}^2$ . The magnetic flux density is 0.5 T. The current in the coil is 2.0 A. Calculate the maximum torque on the coil.
[3]

24

A household ring main circuit is protected by a 30 A circuit breaker. The mains voltage is 240 V. The following appliances are connected to the ring main:

Electric oven: 3.0 kW
Washing machine: 2.2 kW
Microwave: 1.0 kW
Kettle: 2.5 kW
Refrigerator: 0.2 kW

(a) Calculate the total current drawn when all appliances are operating simultaneously.
[2]

(b) Will the circuit breaker trip? Explain your answer.
[2]

(c) The electric oven has a metal casing. Explain why the earth wire is essential for safety.
[2]

(d) The circuit breaker uses an electromagnet. Explain how the circuit breaker trips when the current exceeds 30 A.
[2]

25

A student investigates electromagnetic induction using a coil of wire connected to a centre-zero galvanometer. A bar magnet is moved towards and away from the coil.

(a) State Faraday's Law of Electromagnetic Induction.
[1]

(b) The student observes that the galvanometer deflects to the right when the N-pole of the magnet is moved towards the coil. State the direction of the induced current in the coil (as viewed from the magnet side).
[1]

(c) State Lenz's Law and explain how it applies to this observation.
[2]

(d) Suggest two ways to increase the magnitude of the induced e.m.f.
[2]

(e) The coil has 200 turns and a cross-sectional area of $5.0 \times 10^{-3} \text{ m}^2$ . The magnetic flux density changes from 0.2 T to 0.6 T in 0.5 s. Calculate the average induced e.m.f.
[3]

26

The diagram shows a cathode-ray oscilloscope (CRO) being used to measure the voltage of an a.c. supply. The time-base is set to 2 ms/div and the Y-gain is set to 5 V/div. The trace shows a sinusoidal waveform with a peak-to-peak height of 3.2 divisions and a period of 4.0 divisions.

<image_placeholder> id: Q26-fig1 type: diagram linked_question: Q26 description: CRO screen showing sinusoidal wave. Peak-to-peak = 3.2 divisions. Period = 4.0 divisions. Grid visible. labels: Time-base: 2 ms/div, Y-gain: 5 V/div values: Peak-to-peak = 3.2 div, Period = 4.0 div must_show: Sinusoidal wave, grid divisions, measurements indicated </image_placeholder>

(a) Calculate the peak voltage of the a.c. supply.
[2]

(b) Calculate the root-mean-square (r.m.s.) voltage of the a.c. supply.
[2]

(c) Calculate the frequency of the a.c. supply.
[2]

(d) The time-base is now switched off. Describe the trace seen on the screen.
[1]

Section C: Longer Structured Questions [15 marks]

Answer all questions in the spaces provided.

27

A student designs an experiment to determine the resistivity of a constantan wire. The wire has a length of 1.00 m and a diameter of 0.30 mm. The student measures the resistance of the wire using an ohmmeter and obtains a value of 4.2 $\Omega$ .

(a) Calculate the cross-sectional area of the wire in $\text{m}^2$ .
[2]

(b) Calculate the resistivity of constantan.
[2]

(c) The student repeats the experiment with a wire of the same material but twice the length and half the diameter. Predict the new resistance and explain your reasoning.
[3]

(d) State two precautions the student should take to ensure accurate results.
[2]

28

The diagram shows a model of the national grid electricity transmission system. Electrical energy is generated at 25 kV, stepped up to 400 kV for transmission, and stepped down to 240 V for domestic use.

<image_placeholder> id: Q28-fig1 type: diagram linked_question: Q28 description: National grid model: Generator (25 kV) → Step-up transformer → Transmission lines (400 kV) → Step-down transformer → Consumer (240 V). Transmission line resistance labelled as 10 Ω. labels: 25 kV, 400 kV, 240 V, step-up transformer, step-down transformer, transmission lines, 10 Ω resistance values: 25 kV, 400 kV, 240 V, transmission line resistance = 10 Ω must_show: Complete transmission chain with voltages and line resistance labelled </image_placeholder>

The power station generates 500 MW of electrical power. The total resistance of the transmission lines is 10 $\Omega$ .

(a) Calculate the current in the transmission lines when the voltage is 400 kV.
[2]

(b) Calculate the power loss in the transmission lines.
[2]

(c) Calculate the percentage of generated power lost in transmission.
[2]

(d) Explain why electrical energy is transmitted at high voltage.
[2]

(e) The step-down transformer at the substation has 2000 turns on its primary coil. Calculate the number of turns on its secondary coil.
[2]

End of Paper

Answers

TuitionGoWhere Practice Paper - Pure Physics Secondary 4 (Version 2) - Answer Key

Total Marks: 80

Section A: Multiple Choice Questions [20 marks]

1

Answer: A
Marks: 1
Working:
For a transformer: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{100}{400} = 240 \times 0.25 = 60 \text{ V}$

2

Answer: A
Marks: 1
Explanation:
Use Fleming's Left-Hand Rule (motor rule):

First finger (Field): Left to Right (N to S)
Second finger (Current): Into the page
Thumb (Force): Upwards

3

Answer: A
Marks: 1
Working:
Energy = Power × Time
$= 2.2 \text{ kW} \times \frac{15}{60} \text{ h} = 2.2 \times 0.25 = 0.55 \text{ kWh}$

4

Answer: A
Marks: 1
Working:
Resistivity $\rho = \frac{RA}{l}$
$\rho = \frac{0.34 \times 1.0 \times 10^{-6}}{2.0} = 1.7 \times 10^{-7} \ \Omega \text{m}$

5

Answer: C
Marks: 1
Explanation:
This is Lenz's Law: The direction of the induced current is such that it opposes the change in magnetic flux that produced it.

A is incorrect: e.m.f. is induced when magnet moves away too, or coil moves.
B is incorrect: Induced e.m.f. is directly proportional to rate of change of flux (Faraday's Law).
D is incorrect: Stationary magnet produces no induced current.

6

Answer: B
Marks: 1
Working:
Total resistance = $4 + 2 = 6 \ \Omega$
Circuit current $I = \frac{12}{6} = 2 \text{ A}$
P.d. across $2 \ \Omega$ = $I \times R = 2 \times 2 = 4 \text{ V}$

7

Answer: B
Marks: 1
Explanation:
When the coil is horizontal (parallel to magnetic field), the rate of change of magnetic flux linkage is maximum. The sides of the coil cut magnetic field lines at the maximum rate, so induced e.m.f. is maximum.

8

Answer: D
Marks: 1
Working:
Maximum power = $VI = 240 \times 13 = 3120 \text{ W}$
Number of 60 W lamps = $\frac{3120}{60} = 52$

9

Answer: B
Marks: 1
Explanation:
When the positively charged rod is brought near, electrons in the sphere are attracted to the near side. Earthing allows electrons to flow from earth to the sphere. When the rod is removed, the excess electrons remain, leaving the sphere negatively charged.

10

Answer: B
Marks: 1
Explanation:
The graph shows decreasing gradient (increasing resistance) as voltage increases. For a filament lamp, as voltage increases, the filament gets hotter, and the resistance of the metal filament increases with temperature.

11

Answer: C
Marks: 1
Working:
For an ideal transformer: $\frac{I_s}{I_p} = \frac{N_p}{N_s}$
$I_s = I_p \times \frac{N_p}{N_s} = 0.5 \times \frac{800}{200} = 0.5 \times 4 = 2.0 \text{ A}$

12

Answer: C
Marks: 1
Working:
Period = 4.0 divisions × 2 cycles = 2 divisions per cycle
Time for one cycle = $2 \times 5 \text{ ms} = 10 \text{ ms} = 0.01 \text{ s}$
Frequency $f = \frac{1}{T} = \frac{1}{0.01} = 100 \text{ Hz}$

13

Answer: C
Marks: 1
Working:
Volume constant: $A_1 l_1 = A_2 l_2$
$l_2 = 2l_1 \Rightarrow A_2 = \frac{A_1}{2}$
Resistance $R = \rho \frac{l}{A}$
$R_2 = \rho \frac{2l_1}{A_1/2} = 4 \rho \frac{l_1}{A_1} = 4R$

14

Answer: A
Marks: 1
Explanation:
The fuse must be on the live wire so that when it blows, the appliance is disconnected from the high voltage supply. If the fuse were on the neutral wire, the appliance would remain at live potential even after the fuse blows, posing a shock hazard.

15

Answer: B
Marks: 1
Explanation:
As the magnet falls, it induces eddy currents in the copper tube (Faraday's Law). By Lenz's Law, these currents create a magnetic field that opposes the motion of the magnet, slowing its fall. Lenz's Law specifically explains the opposition to the change.

16

Answer: B
Marks: 1
Working:
Potential divider: $V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$
$V_{out} = 12 \times \frac{3}{6 + 3} = 12 \times \frac{3}{9} = 4 \text{ V}$

17

Answer: B
Marks: 1
Explanation:
A moving charge in a magnetic field experiences a magnetic force $F = Bqv$ (at right angles to both velocity and field). This force acts as the centripetal force causing circular motion.

18

Answer: C
Marks: 1
Working:
Parallel combination: $\frac{1}{R_{total}} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200}$
$R_{total} = \frac{200}{3} \ \Omega$
Total power $P = \frac{V^2}{R_{total}} = \frac{12^2}{200/3} = \frac{144 \times 3}{200} = 2.16 \text{ W}$

19

Answer: A
Marks: 1
Explanation:
Use the right-hand grip rule for solenoids: Curl fingers in direction of current, thumb points to North pole. Based on the current direction shown in the diagram (which would be specified in the actual diagram), End P is the North pole.

20

Answer: (Graph showing exponential rise)
Marks: 1
Explanation:
When a capacitor charges through a resistor, the voltage across it rises exponentially: $V = V_0(1 - e^{-t/RC})$ . It starts at 0 and asymptotically approaches the supply voltage.

Section B: Structured Questions [45 marks]

21

(a) [3 marks]
Circuit diagram should show:

Battery symbol (long line +, short line -)
Variable resistor (rectangle with diagonal arrow)
Ammeter (circle with A) in series
Voltmeter (circle with V) in parallel across the conductor
Conductor (rectangle or wire symbol)
All components correctly connected in series/parallel as appropriate
Labels for each component

Mark breakdown:

Correct symbols for all 5 components: 1 mark
Correct series/parallel connections: 1 mark
Proper labelling: 1 mark

(b) [3 marks]
Graph plotting:

Axes labelled with units (Voltage/V, Current/A): 1 mark
Suitable scales (using >50% of grid): 1 mark
All 6 points plotted correctly (± half a small square): 1 mark
Best-fit straight line through origin: (included in plotting mark)

(c) [2 marks]
Resistance determination:

Gradient of graph = $\frac{\Delta I}{\Delta V} = \frac{1.2 - 0}{3.0 - 0} = 0.4 \text{ A/V}$
Resistance $R = \frac{1}{\text{gradient}} = \frac{1}{0.4} = 2.5 \ \Omega$
OR using any point: $R = \frac{V}{I} = \frac{3.0}{1.2} = 2.5 \ \Omega$

Mark breakdown:

Correct method (gradient or V/I): 1 mark
Correct answer with unit: 1 mark

(d) [1 mark]
Answer: Ohm's Law
Explanation: The conductor obeys Ohm's Law (current is directly proportional to voltage at constant temperature), as shown by the straight-line graph through the origin.

22

(a) [2 marks]
Working:
$\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$N_s = N_p \times \frac{V_s}{V_p} = 1200 \times \frac{12}{240} = 1200 \times 0.05 = 60 \text{ turns}$

Mark breakdown:

Correct formula/substitution: 1 mark
Correct answer: 1 mark

(b) [3 marks]
Working:
Efficiency $\eta = \frac{P_{out}}{P_{in}} = \frac{V_s I_s}{V_p I_p} = 0.90$
$I_p = \frac{V_s I_s}{\eta V_p} = \frac{12 \times 4.0}{0.90 \times 240} = \frac{48}{216} = 0.222 \text{ A}$

Mark breakdown:

Correct efficiency formula: 1 mark
Correct substitution: 1 mark
Correct answer with unit: 1 mark

Common mistake: Forgetting to convert 90% to 0.90, giving $I_p = 0.0222 \text{ A}$ (incorrect).

(c) [2 marks]
Answer:
The core is laminated to reduce eddy currents. The insulating layers between laminations increase the resistance to eddy current flow, reducing energy loss as heat.

Mark breakdown:

Mention eddy currents: 1 mark
Explain how lamination reduces them (increases resistance/path broken): 1 mark

(d) [1 mark]
Answer (any one):

Resistance of coils (copper losses / $I^2R$ losses)
Eddy current losses in core
Hysteresis losses in core
Flux leakage (not all flux links both coils)

23

(a) [2 marks]
On diagram:

Force on AB: Upwards (use Fleming's Left-Hand Rule: Field left→right, Current A→B, Force up)
Force on CD: Downwards (Current C→D opposite to AB, so force opposite)

Mark breakdown:

Correct direction on AB: 1 mark
Correct direction on CD: 1 mark

(b) [2 marks]
Explanation:
The forces on AB and CD are equal in magnitude, opposite in direction, and not along the same line of action. They form a couple which produces a turning effect (torque) on the coil.

Mark breakdown:

Forces are equal, opposite, parallel but not collinear: 1 mark
Form a couple / produce torque: 1 mark

(c) [1 mark]
Answer:
The commutator reverses the current direction in the coil every half-turn, ensuring the torque always acts in the same direction so the coil rotates continuously.

(d) [3 marks]
Working:
Maximum torque $\tau_{max} = B I A N$ (when coil is parallel to field)
$A = 4.0 \times 10^{-3} \text{ m}^2$ (area per turn)
$\tau_{max} = 0.5 \times 2.0 \times (4.0 \times 10^{-3}) \times 50$
$= 0.5 \times 2.0 \times 0.2 = 0.2 \text{ N m}$

Mark breakdown:

Correct formula: 1 mark
Correct substitution: 1 mark
Correct answer with unit: 1 mark

24

(a) [2 marks]
Working:
Total power = $3.0 + 2.2 + 1.0 + 2.5 + 0.2 = 8.9 \text{ kW} = 8900 \text{ W}$
Total current $I = \frac{P}{V} = \frac{8900}{240} = 37.1 \text{ A}$

Mark breakdown:

Correct total power: 1 mark
Correct current calculation with unit: 1 mark

(b) [2 marks]
Answer:
Yes, the circuit breaker will trip. The total current (37.1 A) exceeds the circuit breaker rating (30 A).

Mark breakdown:

Correct yes/no: 1 mark
Correct reasoning with comparison: 1 mark

(c) [2 marks]
Explanation:
If the live wire touches the metal casing, the casing becomes live. The earth wire provides a low-resistance path for current to flow to ground, causing a large current that blows the fuse / trips the circuit breaker, disconnecting the supply and preventing electric shock.

Mark breakdown:

Earth wire provides path to ground: 1 mark
Causes fuse to blow / breaker to trip / prevents shock: 1 mark

(d) [2 marks]
Explanation:
When current exceeds 30 A, the electromagnet becomes strong enough to attract the iron armature. This releases the latch, opening the contacts and breaking the circuit.

Mark breakdown:

Electromagnet attracts armature: 1 mark
Contacts open / circuit breaks: 1 mark

25

(a) [1 mark]
Answer:
The magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage.
OR: $E = -\frac{d(N\Phi)}{dt}$

(b) [1 mark]
Answer:
Anticlockwise (as viewed from the magnet side).

Explanation:
N-pole approaching → flux into coil increases → induced current creates N-pole facing the magnet to repel it (Lenz's Law) → right-hand grip rule gives anticlockwise current when viewed from magnet side.

(c) [2 marks]
Lenz's Law: The direction of the induced current is such that it opposes the change in magnetic flux that produces it.

Application: As the N-pole approaches, magnetic flux through the coil increases. The induced current flows to create a magnetic field opposing this increase (i.e., an N-pole facing the approaching magnet), which repels the magnet. This opposition is the manifestation of Lenz's Law.

Mark breakdown:

Statement of Lenz's Law: 1 mark
Application to observation: 1 mark

(d) [2 marks]
Any two:

Increase the number of turns on the coil
Increase the speed of the magnet
Use a stronger magnet (increase magnetic flux density)
Increase the cross-sectional area of the coil

Mark breakdown:

1 mark per valid suggestion

<stage3_exam_answers_md>

TuitionGoWhere Practice Paper - Pure Physics Secondary 4

TuitionGoWhere Secondary School (AI)
Subject: Pure Physics
Level: Secondary 4
Paper: PRELIM (Version 2)
Duration: 1 hour 45 minutes
Total Marks: 80

Answer Key and Marking Scheme

Section A: Multiple Choice Questions [20 marks]

Question	Answer	Explanation
1	A	$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{100}{400} = 60 \text{ V}$
2	A	Fleming's Left Hand Rule: Field (Left→Right), Current (Into page) → Force (Upwards)
3	A	Energy = Power × Time = $2.2 \text{ kW} \times \frac{15}{60} \text{ h} = 0.55 \text{ kWh}$
4	A	$\rho = \frac{RA}{L} = \frac{0.34 \times 1.0 \times 10^{-6}}{2.0} = 1.7 \times 10^{-7} \ \Omega \text{m}$
5	C	Lenz's Law: Induced current opposes the change producing it
6	B	Total R = 6 Ω, I = 12/6 = 2 A, $V_{2\Omega} = 2 \times 2 = 4 \text{ V}$
7	B	Coil horizontal → rate of change of flux is maximum → induced e.m.f. maximum
8	D	Max power = $240 \times 13 = 3120 \text{ W}$ , Number = $3120 / 60 = 52$
9	B	Earthing with +ve rod nearby → electrons flow from earth → sphere becomes negative
10	B	Filament lamp: resistance increases with temperature → gradient decreases
11	C	$I_s = I_p \times \frac{N_p}{N_s} = 0.5 \times \frac{800}{200} = 2.0 \text{ A}$
12	C	Period = $4 \text{ div} \times 5 \text{ ms/div} = 20 \text{ ms}$ , $f = 1/0.02 = 50 \text{ Hz}$
13	C	Length doubles, area halves (constant volume) → $R_{new} = \rho \frac{2L}{A/2} = 4R$
14	A	Fuse is always on the live wire
15	B	Eddy currents induced (Faraday), opposing magnet's motion (Lenz) → magnetic braking
16	B	$V_{out} = 12 \times \frac{3}{6+3} = 4 \text{ V}$
17	B	Magnetic force $F = Bqv$ provides centripetal force
18	C	$P = \frac{V^2}{R_{eq}}$ , $R_{eq} = \frac{100 \times 200}{300} = 66.7 \Omega$ , $P = \frac{144}{66.7} = 2.16 \text{ W}$
19	A	Right-hand grip rule: fingers curl in current direction, thumb points to North pole
20	Correct graph: Exponential rise	$V_C = V_0(1 - e^{-t/RC})$

Section B: Structured Questions [45 marks]

21

(a) [3 marks]

Battery symbol correct (long/short lines) [1]
Variable resistor symbol correct (rectangle with diagonal arrow) [1]
Ammeter in series, voltmeter in parallel across conductor, all labels present [1]

(b) [3 marks]

Axes labelled with units (V and A) [1]
Appropriate scale covering >50% grid, points plotted correctly (±½ small square) [1]
Best-fit straight line through origin [1]

(c) [2 marks]

Gradient calculation shown: $\frac{\Delta I}{\Delta V} = \frac{1.2 - 0}{3.0 - 0} = 0.4 \ \text{A/V}$ [1]
Resistance = $1/\text{gradient} = 2.5 \ \Omega$ [1]

(d) [1 mark]

Ohm's Law [1]

22

(a) [2 marks]

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$ → $N_s = 1200 \times \frac{12}{240} = 60$ turns [2]

(b) [3 marks]

$P_{out} = V_s I_s = 12 \times 4.0 = 48 \text{ W}$ [1]
$P_{in} = \frac{P_{out}}{0.90} = \frac{48}{0.9} = 53.33 \text{ W}$ [1]
$I_p = \frac{P_{in}}{V_p} = \frac{53.33}{240} = 0.222 \text{ A}$ [1]

(c) [2 marks]

Laminations reduce eddy currents [1]
Eddy currents cause heating/energy loss in core [1]

(d) [1 mark]

Any one: Resistance of windings (copper losses), hysteresis losses, flux leakage, eddy currents (if not in c)

23

(a) [2 marks]

Force on AB: Downwards (Fleming's LHR: Field left→right, Current A→B) [1]
Force on CD: Upwards (Field left→right, Current D→C) [1]

(b) [2 marks]

Forces on AB and CD are equal, opposite, and not collinear [1]
They form a couple producing a clockwise turning moment [1]

(c) [1 mark]

Reverses current direction in coil every half-turn to maintain continuous rotation in same direction [1]

(d) [3 marks]

$\tau_{max} = N B I A = 50 \times 0.5 \times 2.0 \times 4.0 \times 10^{-3}$ [1]
$= 0.20 \ \text{N m}$ [1]
Unit correct [1]

24

(a) [2 marks]

Total power = $3.0 + 2.2 + 1.0 + 2.5 + 0.2 = 8.9 \text{ kW}$ [1]
$I = \frac{P}{V} = \frac{8900}{240} = 37.1 \text{ A}$ [1]

(b) [2 marks]

Yes, circuit breaker will trip [1]
Total current (37.1 A) > 30 A rating [1]

(c) [2 marks]

If live wire touches casing, earth wire provides low-resistance path to ground [1]
Large current flows, blowing fuse/tripping breaker, preventing electric shock [1]

(d) [2 marks]

Current > 30 A → electromagnet strong enough to attract iron latch [1]
Latch releases, spring opens contacts, breaking circuit [1]

25

(a) [1 mark]

Induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage [1]

(b) [1 mark]

Anticlockwise (as viewed from magnet side) [1]

(c) [2 marks]

Lenz's Law: Induced current flows in direction to oppose the change producing it [1]
N-pole approaching → coil top becomes N-pole to repel magnet → anticlockwise current [1]

(d) [2 marks]

Any two: Increase magnet speed, increase magnet strength, increase number of turns, increase coil area [2]

(e) [3 marks]

$\Delta \Phi = N A \Delta B = 200 \times 5.0 \times 10^{-3} \times (0.6 - 0.2) = 0.4 \ \text{Wb}$ [1]
$\mathcal{E} = \frac{\Delta \Phi}{\Delta t} = \frac{0.4}{0.5} = 0.8 \ \text{V}$ [1]
Unit correct [1]

26

(a) [2 marks]

Peak-to-peak = $3.2 \times 5 = 16 \text{ V}$ [1]
Peak voltage = $16 / 2 = 8 \text{ V}$ [1]

(b) [2 marks]

$V_{rms} = \frac{V_{peak}}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 5.66 \text{ V}$ [2] (1 for formula, 1 for answer)

(c) [2 marks]

Period = $4.0 \times 2 \text{ ms} = 8 \text{ ms}$ [1]
$f = \frac{1}{T} = \frac{1}{8 \times 10^{-3}} = 125 \text{ Hz}$ [1]

(d) [1 mark]

Vertical line (length = peak-to-peak = 3.2 divisions) [1]

Section C: Longer Structured Questions [15 marks]

27

(a) [2 marks]

Radius = $0.15 \text{ mm} = 1.5 \times 10^{-4} \text{ m}$ [1]
$A = \pi r^2 = \pi \times (1.5 \times 10^{-4})^2 = 7.07 \times 10^{-8} \text{ m}^2$ [1]

(b) [2 marks]

$\rho = \frac{RA}{L} = \frac{4.2 \times 7.07 \times 10^{-8}}{1.00} = 2.97 \times 10^{-7} \ \Omega \text{m}$ [2] (1 for substitution, 1 for answer)

(c) [3 marks]

New length = 2.00 m, new diameter = 0.15 mm → new area = $\frac{1}{4}$ original [1]
$R_{new} = \rho \frac{2L}{A/4} = 8 \times \rho \frac{L}{A} = 8 \times 4.2 = 33.6 \ \Omega$ [1]
Explanation: R ∝ L, R ∝ 1/A, A ∝ d² → doubling L and halving d gives 8× resistance [1]

(d) [2 marks]

Any two: Avoid heating (use low current/switch off between readings), measure diameter in several places with micrometer, ensure wire is straight and taut, use 4-terminal measurement for low resistances [2]

28

(a) [2 marks]

High voltage → low current for same power ( $P = VI$ ) [1]
Low current → reduced $I^2R$ power loss in transmission cables [1]

(b) [2 marks]

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$ → $\frac{400,000}{25,000} = \frac{N_s}{N_p} = 16$ [1]
Turns ratio = 16:1 [1]

(c) [3 marks]

$P_{loss} = I^2 R$ [1]
At 25 kV: $I = \frac{P}{25,000}$ , $P_{loss} = \left(\frac{P}{25,000}\right)^2 R$ [1]
At 400 kV: $I = \frac{P}{400,000}$ , $P_{loss} = \left(\frac{P}{400,000}\right)^2 R = \frac{1}{256} \times$ loss at 25 kV [1]

(d) [2 marks]

Any two: Insulation breakdown at very high voltages, corona discharge losses, increased cost of insulation/clearances, safety hazards, difficulty in switching/protection [2]

(e) [2 marks]

Step-down transformers reduce voltage to safer levels for domestic use [1]
High voltage (400 kV) is dangerous and unsuitable for household appliances [1]

End of Marking Scheme Total: 80 marks